IIT Madras
Mathematics 2
Graded Assignment 5

Graded Assignment 5

Exercise Questions ❓

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Exercise Solutions 🔬

Hello! I can certainly help you with these linear algebra questions. Here is a detailed breakdown of each problem with the core concepts and step-by-step solutions.

❓ Question 1: Student Marks

The marks obtained by Karthika, Romy and Farzana in Quiz 1, Quiz 2 and End sem (with the maximum marks for each exam being 100) are shown in Table M2W5G1.

Quiz 1Quiz 2End sem
Karthika405060
Romy205050
Farzana304070

If $x_1%$ of Quiz 1, $x_2%$ of Quiz 2, and $x_3%$ of End sem is taken to calculate the total marks $T_1(x_1, x_2, x_3)$, $T_2(x_1, x_2, x_3)$, and $T_3(x_1, x_2, x_3)$ of Karthika, Romy, and Farzana, respectively, then choose the correct set of options.

  • $T_2(x_1, x_2, x_3) = \frac{1}{100}(20x_1 + 50x_2 + 50x_3)$
  • $T_2(x_1, x_2, x_3) = \frac{1}{100}(40x_1 + 20x_2 + 30x_3)$
  • If $x_1 = 20%$, $x_2 = 20%$, and $x_3 = 60%$, then $T_1(x_1, x_2, x_3) = 44$.
  • If $x_1 = 20%$, $x_2 = 20%$, and $x_3 = 60%$, then $T_1(x_1, x_2, x_3) = 54$.
  • If $x_1 = 20%$, $x_2 = 20%$, and $x_3 = 60%$, then Farzana obtained the highest total marks.
  • $T_i(x_1, x_2, x_3)$ is a linear mapping from $\mathbb{R}^3$ to $\mathbb{R}$, for each $i = 1, 2, 3$.

Core Concepts

  1. Weighted Average: The total score is a weighted sum of the individual scores. If the scores are $s_1, s_2, s_3$ and the weights are $x_1%, x_2%, x_3%$, the total $T$ is: $$T = s_1 \left(\frac{x_1}{100}\right) + s_2 \left(\frac{x_2}{100}\right) + s_3 \left(\frac{x_3}{100}\right) = \frac{1}{100}(s_1 x_1 + s_2 x_2 + s_3 x_3)$$
  2. Linear Transformation: A mapping $T: \mathbb{R}^n \to \mathbb{R}^m$ is linear if $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ and $T(c\mathbf{u}) = cT(\mathbf{u})$. A mapping from $\mathbb{R}^3$ to $\mathbb{R}$ of the form $T(x_1, x_2, x_3) = c_1 x_1 + c_2 x_2 + c_3 x_3$ (a linear combination) is a linear mapping.

Solution

Let’s write the functions for each student using the formula:

  • Karthika ($T_1$): Scores [40, 50, 60] $\implies T_1 = \frac{1}{100}(40x_1 + 50x_2 + 60x_3)$
  • Romy ($T_2$): Scores [20, 50, 50] $\implies T_2 = \frac{1}{100}(20x_1 + 50x_2 + 50x_3)$
  • Farzana ($T_3$): Scores [30, 40, 70] $\implies T_3 = \frac{1}{100}(30x_1 + 40x_2 + 70x_3)$

Now, let’s check each option:

  • $T_2(x_1, x_2, x_3) = \frac{1}{100}(20x_1 + 50x_2 + 50x_3)$
    • This perfectly matches our derived formula for Romy ($T_2$). This is correct.
  • $T_2(x_1, x_2, x_3) = \frac{1}{100}(40x_1 + 20x_2 + 30x_3)$
    • This does not match the formula for $T_2$. This is incorrect.
  • If $x_1 = 20, x_2 = 20, x_3 = 60$, then $T_1 = 44$.
    • $T_1 = \frac{1}{100}(40(20) + 50(20) + 60(60))$
    • $T_1 = \frac{1}{100}(800 + 1000 + 3600) = \frac{1}{100}(5400) = 54$.
    • The option says 44. This is incorrect.
  • If $x_1 = 20, x_2 = 20, x_3 = 60$, then $T_1 = 54$.
    • As calculated above, $T_1 = 54$. This is correct.
  • If $x_1 = 20, x_2 = 20, x_3 = 60$, then Farzana obtained the highest total marks.
    • We need to calculate all three scores:
    • $T_1$ (Karthika) = 54 (from above)
    • $T_2$ (Romy) = $\frac{1}{100}(20(20) + 50(20) + 50(60)) = \frac{1}{100}(400 + 1000 + 3000) = 44$
    • $T_3$ (Farzana) = $\frac{1}{100}(30(20) + 40(20) + 70(60)) = \frac{1}{100}(600 + 800 + 4200) = 56$
    • The scores are 54, 44, and 56. Farzana’s score (56) is the highest. This is correct.
  • $T_i(x_1, x_2, x_3)$ is a linear mapping from $\mathbb{R}^3$ to $\mathbb{R}$, for each $i = 1, 2, 3$.
    • Yes. Each function $T_i$ is of the form $c_1 x_1 + c_2 x_2 + c_3 x_3$ (where the $c$’s are the scores divided by 100). This is the definition of a linear mapping from $\mathbb{R}^3$ to $\mathbb{R}$. This is correct.

✅ Correct Options:

  • $T_2(x_1, x_2, x_3) = \frac{1}{100}(20x_1 + 50x_2 + 50x_3)$
  • If $x_1 = 20%$, $x_2 = 20%$, and $x_3 = 60%$, then $T_1(x_1, x_2, x_3) = 54$.
  • If $x_1 = 20%$, $x_2 = 20%$, and $x_3 = 60%$, then Farzana obtained the highest total marks.
  • $T_i(x_1, x_2, x_3)$ is a linear mapping from $\mathbb{R}^3$ to $\mathbb{R}$, for each $i = 1, 2, 3$.

❓ Question 2: Balancing Chemical Equations

…The chemical equation corresponding to this chemical reaction (R) is given below. $x_1 CO_2 + x_2 H_2O \to x_3 C_6H_{12}O_6 + x_4 O_2$ …(R) … Consider the system of linear equations obtained for balancing the chemical equation (R) and choose the set of correct options.

  • The nullity of the matrix corresponding to this system is 1.
  • The nullity of the matrix corresponding to this system is 2.
  • ${(6, 6, 1, 6)}$ is a basis of the null space of the matrix corresponding to this system.
  • ${(1, 1, 6, 1)}$ is a basis of the null space of the matrix corresponding to this system.
  • ${(1, 1, 1, 6), (6, 6, 1, 1)}$ is a basis of the null space of the matrix corresponding to this system.
  • ${(6, 6, 0, 6), (0, 0, 1, 0)}$ is a basis of the null space of the matrix corresponding to this system.
  • There are infinitely many ways to balance the chemical equation (R).

Core Concepts

  1. Balancing Equations: To balance a chemical equation, you conserve the number of atoms of each element on both sides. This creates a system of linear equations.
  2. Homogeneous System: The resulting system is “homogeneous” (of the form $A\mathbf{x} = \mathbf{0}$) because all constants are zero.
  3. Null Space: The set of all solutions $\mathbf{x}$ to $A\mathbf{x} = \mathbf{0}$ is called the null space of $A$.
  4. Nullity: The dimension of the null space (the number of free variables, or the number of vectors in a basis) is the nullity.

Solution

Let’s build the system of equations by balancing each atom type:

  • Carbon (C): $x_1 = 6x_3 \implies x_1 - 6x_3 = 0$
  • Hydrogen (H): $2x_2 = 12x_3 \implies x_2 - 6x_3 = 0$
  • Oxygen (O): $2x_1 + x_2 = 6x_3 + 2x_4 \implies 2x_1 + x_2 - 6x_3 - 2x_4 = 0$

This is our system $A\mathbf{x} = \mathbf{0}$. Let’s solve it.

  1. From the first two equations, we see that $x_1$ and $x_2$ depend on $x_3$. Let’s choose $x_3$ as a free variable.
    • Let $x_3 = t$
    • From $x_1 - 6x_3 = 0 \implies x_1 = 6x_3 = 6t$
    • From $x_2 - 6x_3 = 0 \implies x_2 = 6x_3 = 6t$
  2. Now substitute these into the third equation to find $x_4$:
    • $2(6t) + (6t) - 6(t) - 2x_4 = 0$
    • $12t + 6t - 6t - 2x_4 = 0$
    • $12t - 2x_4 = 0 \implies 2x_4 = 12t \implies x_4 = 6t$
  3. The solution vector $\mathbf{x} = (x_1, x_2, x_3, x_4)$ is:
    • $\mathbf{x} = (6t, 6t, t, 6t)$
    • We can factor out the free variable $t$: $\mathbf{x} = t(6, 6, 1, 6)$

This means the null space is spanned by the single vector $(6, 6, 1, 6)$.

  • The basis for the null space is ${(6, 6, 1, 6)}$.
  • The nullity (dimension of the basis) is 1.
  • Since $t$ can be any real number, there are infinitely many mathematical solutions.

Let’s check the options:

  • The nullity… is 1. Correct.
  • The nullity… is 2. Incorrect.
  • ${(6, 6, 1, 6)}$ is a basis… Correct.
  • ${(1, 1, 6, 1)}$ is a basis… Incorrect.
  • ${(1, 1, 1, 6), (6, 6, 1, 1)}$ is a basis… Incorrect.
  • ${(6, 6, 0, 6), (0, 0, 1, 0)}$ is a basis… Incorrect.
  • There are infinitely many ways to balance… Correct. The solution set is a line in $\mathbb{R}^4$, containing infinite vectors.

✅ Correct Options:

  • The nullity of the matrix corresponding to this system is 1.
  • ${(6, 6, 1, 6)}$ is a basis of the null space of the matrix corresponding to this system.
  • There are infinitely many ways to balance the chemical equation (R).

❓ Question 3: Scalar Matrix

Choose the set of correct options.

  • The null-space of a scalar matrix is always the zero vector space.
  • The null-space of a non-zero scalar matrix is always the zero vector space.
  • The nullity of a non-zero scalar matrix is always 0.
  • The rank of a non-zero scalar matrix is always same as the order of the matrix.
  • The rank of a non-zero scalar matrix is always 0.

Core Concepts

  1. Scalar Matrix: An $n \times n$ matrix $A$ of the form $A = kI$, where $I$ is the identity matrix and $k$ is a scalar. $$A = \begin{pmatrix} k & 0 & \dots \\ 0 & k & \dots \\ \vdots & \vdots & \ddots \end{pmatrix}$$
  2. Zero Matrix: If $k=0$, the scalar matrix is the zero matrix ($\mathbf{0}$).
  3. Non-zero Scalar Matrix: If $k \ne 0$.
  4. Null Space ($A\mathbf{x} = \mathbf{0}$): The set of solutions $\mathbf{x}$.
  5. Nullity: The dimension of the null space.
  6. Rank: The number of linearly independent rows/columns.
  7. Rank-Nullity Theorem: For an $n \times n$ matrix, $\text{rank}(A) + \text{nullity}(A) = n$.

Solution

Let’s check each option:

  • The null-space of a scalar matrix is always the zero vector space.
    • This is false. Consider the scalar matrix with $k=0$ (the zero matrix). The equation $A\mathbf{x} = \mathbf{0}$ becomes $\mathbf{0}\mathbf{x} = \mathbf{0}$, which is true for any vector $\mathbf{x}$. The null space is the entire $\mathbb{R}^n$, not just ${\mathbf{0}}$.
  • The null-space of a non-zero scalar matrix is always the zero vector space.
    • “Non-zero” means $k \ne 0$. We are solving $A\mathbf{x} = \mathbf{0}$, or $(kI)\mathbf{x} = \mathbf{0}$.
    • This simplifies to $k(I\mathbf{x}) = \mathbf{0}$, or $k\mathbf{x} = \mathbf{0}$.
    • Since $k \ne 0$, we can divide by it, leaving $\mathbf{x} = \mathbf{0}$.
    • The only solution is the trivial solution. The null space is ${\mathbf{0}}$, which is the zero vector space. This is correct.
  • The nullity of a non-zero scalar matrix is always 0.
    • From the previous point, the null space is ${\mathbf{0}}$. The dimension of the zero vector space is 0.
    • Therefore, the nullity is 0. This is correct.
  • The rank of a non-zero scalar matrix is always same as the order of the matrix.
    • Let the matrix be $n \times n$ (order $n$). Since it’s a non-zero scalar matrix, $k \ne 0$.
    • The matrix is $\text{diag}(k, k, \dots, k)$. The $n$ columns are linearly independent.
    • Thus, the rank is $n$.
    • Alternatively, using Rank-Nullity: $\text{rank}(A) + \text{nullity}(A) = n$. From the previous point, $\text{nullity}(A) = 0$. So, $\text{rank}(A) + 0 = n \implies \text{rank}(A) = n$. This is correct.
  • The rank of a non-zero scalar matrix is always 0.
    • This is false. As just shown, the rank is $n$.

✅ Correct Options:

  • The null-space of a non-zero scalar matrix is always the zero vector space.
  • The nullity of a non-zero scalar matrix is always 0.
  • The rank of a non-zero scalar matrix is always same as the order of the matrix.

❓ Question 4: $A \in M_{5 \times 3}(\mathbb{R})$

Let $A \in M_{5 \times 3}(\mathbb{R})$ be a non-zero matrix. Select all true statements from the following:

  • The nullity of $A$ can be at most 2.
  • The nullity of $A$ has to be at least 2.
  • The rank of $A$ can be at most 3.
  • The rank of $A$ has to be at least 3.

Core Concepts

  1. Matrix Dimensions: $A$ is a $5 \times 3$ matrix.
    • $m = 5$ (rows, dimension of the target space $\mathbb{R}^5$)
    • $n = 3$ (columns, dimension of the domain $\mathbb{R}^3$)
  2. Rank Bounds: The rank is the number of linearly independent columns (or rows).
    • The number of independent columns can’t be more than the number of columns ($n=3$).
    • The number of independent rows can’t be more than the number of rows ($m=5$).
    • Therefore, $\text{rank}(A) \le \min(m, n) = \min(5, 3) = 3$.
  3. Rank-Nullity Theorem: $\text{rank}(A) + \text{nullity}(A) = n$ (number of columns).
    • For this matrix, $\text{rank}(A) + \text{nullity}(A) = 3$.

Solution

  1. Bounds on Rank:
    • We know $\text{rank}(A) \le 3$.
    • The problem states $A$ is a non-zero matrix, which means its rank must be at least 1. (A rank of 0 implies it’s the zero matrix).
    • So, the true range for rank is: $1 \le \text{rank}(A) \le 3$.
  2. Bounds on Nullity:
    • From the Rank-Nullity Theorem: $\text{nullity}(A) = 3 - \text{rank}(A)$.
    • Let’s use the range of rank to find the range of nullity:
      • If $\text{rank}(A) = 1$ (its minimum), $\text{nullity}(A) = 3 - 1 = 2$.
      • If $\text{rank}(A) = 3$ (its maximum), $\text{nullity}(A) = 3 - 3 = 0$.
    • So, the true range for nullity is: $0 \le \text{nullity}(A) \le 2$.

Now, let’s check the options:

  • The nullity of $A$ can be at most 2.
    • Our range is $0 \le \text{nullity}(A) \le 2$. The maximum value is 2. This is correct.
  • The nullity of $A$ has to be at least 2.
    • This is false. If $\text{rank}(A) = 3$, the nullity would be 0.
  • The rank of $A$ can be at most 3.
    • Our range is $1 \le \text{rank}(A) \le 3$. The maximum value is 3. This is correct.
  • The rank of $A$ has to be at least 3.
    • This is false. If $A$ has only one non-zero column, the rank could be 1.

✅ Correct Options:

  • The nullity of $A$ can be at most 2.
  • The rank of $A$ can be at most 3.

❓ Question 5: Linear Transformations

Which of the following maps are linear transformations?

  • $T : \mathbb{R}^2 \to \mathbb{R}^3$ defined by $T(x, y) = (2x + 1, 7x, 7x + 10y + 6)$.
  • $T : \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(x, y) = (7x + 4y, 2x - 10y)$.
  • $T : \mathbb{R}^2 \to \mathbb{R}$ defined by $T(x, y) = 7x - 7y$.
  • $T : \mathbb{R} \to \mathbb{R}^2$ defined by $T(x) = (10|x|, 0)$.

Core Concepts

A transformation $T$ is linear if it satisfies two conditions:

  1. Additivity: $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ for all $\mathbf{u}, \mathbf{v}$.
  2. Homogeneity: $T(c\mathbf{u}) = cT(\mathbf{u})$ for all scalars $c$ and vectors $\mathbf{u}$.

A simple test:

  1. It must map the zero vector to the zero vector: $T(\mathbf{0}) = \mathbf{0}$. If not, it’s not linear.
  2. Each output component must be a linear combination of the input variables (e.g., $ax + by$). No constants, powers ($x^2$), or non-linear functions ($|x|, \sin(x), \sqrt{x}$).

Solution

  • $T(x, y) = (2x + 1, 7x, 7x + 10y + 6)$
    • Test $T(\mathbf{0})$: $T(0, 0) = (2(0) + 1, 7(0), 7(0) + 10(0) + 6) = (1, 0, 6)$.
    • This is not the zero vector $(0, 0, 0)$. It is not linear.
  • $T(x, y) = (7x + 4y, 2x - 10y)$
    • Test $T(\mathbf{0})$: $T(0, 0) = (0, 0)$. This passes.
    • Check components:
      • 1st component: $7x + 4y$. This is a linear combination.
      • 2nd component: $2x - 10y$. This is a linear combination.
    • This transformation is linear.
  • $T(x, y) = 7x - 7y$
    • Test $T(\mathbf{0})$: $T(0, 0) = 0$. This passes.
    • Check components: The only component is $7x - 7y$, which is a linear combination.
    • This transformation is linear.
  • $T(x) = (10|x|, 0)$
    • Test $T(\mathbf{0})$: $T(0) = (10|0|, 0) = (0, 0)$. This passes.
    • Let’s test homogeneity: $T(c x) = c T(x)$.
    • Let $x = 1$ and $c = -1$.
    • $T(c x) = T(-1 \cdot 1) = T(-1) = (10|-1|, 0) = (10, 0)$.
    • $c T(x) = -1 \cdot T(1) = -1 \cdot (10|1|, 0) = -1 \cdot (10, 0) = (-10, 0)$.
    • Since $(10, 0) \ne (-10, 0)$, $T(cx) \ne cT(x)$. It is not linear.

✅ Correct Options:

  • $T : \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(x, y) = (7x + 4y, 2x - 10y)$.
  • $T : \mathbb{R}^2 \to \mathbb{R}$ defined by $T(x, y) = 7x - 7y$.

❓ Question 6: Map $T(v) = Av$

Consider a map $T : \mathbb{R}^2 \to \mathbb{R}^2$ defined as $T(v) = Av$, where $v = \begin{bmatrix} x \ y \end{bmatrix}$ and $A = \begin{bmatrix} -18 & -11 \ 0 & -15 \end{bmatrix}$. Which of the following options are correct?

  • $T$ is both one-one and onto.
  • $T$ is a linear transformation.
  • $T$ is one-one but not onto.
  • $T$ is neither one-one nor onto.

Core Concepts

  1. Matrix Transformation: Any transformation defined by $T(\mathbf{v}) = A\mathbf{v}$ (matrix multiplication) is always a linear transformation.
  2. Invertible Matrix Theorem (IMT): For a square $n \times n$ matrix $A$, the following statements are equivalent (all true or all false):
    • $A$ is invertible.
    • $\det(A) \ne 0$.
    • The transformation $T(\mathbf{v}) = A\mathbf{v}$ is one-to-one (injective).
    • The transformation $T(\mathbf{v}) = A\mathbf{v}$ is onto (surjective).

Solution

  1. Check for Linearity:
    • The map is defined as $T(v) = Av$. By definition, any transformation defined by matrix multiplication is a linear transformation.
    • Therefore, "$T$ is a linear transformation" is correct.
  2. Check for One-to-one and Onto:
    • Since $A$ is a square ($2 \times 2$) matrix, we can use the determinant to check its properties.
    • $A$ is an upper triangular matrix (all entries below the main diagonal are zero).
    • The determinant of a triangular matrix is the product of its diagonal entries.
    • $\det(A) = (-18) \cdot (-15) = 270$.
    • Since $\det(A) = 270 \ne 0$, the matrix $A$ is invertible.
    • According to the Invertible Matrix Theorem, the transformation $T$ is both one-to-one and onto.
    • Therefore, "$T$ is both one-one and onto" is correct.

Let’s check the options:

  • $T$ is both one-one and onto. Correct.
  • $T$ is a linear transformation. Correct.
  • $T$ is one-one but not onto. Incorrect.
  • $T$ is neither one-one nor onto. Incorrect.

✅ Correct Options:

  • $T$ is both one-one and onto.
  • $T$ is a linear transformation.

❓ Question 7: Basis for Null Space

Consider the matrix $A$ given as follows: $A= \begin{bmatrix} 3 & 2 & 0 & 0 \ 0 & 5 & 0 & 0 \ 1 & 0 & 3 & 0 \ 0 & 0 & 0 & 0 \end{bmatrix}$ Which of the following sets is a possible basis for the null space of $A$?

  • ${(0, 0, 1, 0), (0, 0, 0, 1)}$
  • ${(0, 1, 0, 0), (0, 0, 0, 1)}$
  • ${(0, 0, 1, 0)}$
  • ${(0, 0, 0, 1)}$

Core Concepts

The null space of a matrix $A$ is the set of all vectors $\mathbf{x}$ that solve the homogeneous equation $A\mathbf{x} = \mathbf{0}$. A basis for the null space is a set of linearly independent vectors that span this solution set.

Solution

We need to solve $A\mathbf{x} = \mathbf{0}$, where $\mathbf{x} = \begin{pmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{pmatrix}$. This gives the following system of equations:

  1. $3x_1 + 2x_2 = 0$
  2. $5x_2 = 0$
  3. $x_1 + 3x_3 = 0$
  4. $0 = 0$

Let’s solve this system from the simplest equation:

  • From (2): $5x_2 = 0 \implies x_2 = 0$.
  • Substitute $x_2 = 0$ into (1): $3x_1 + 2(0) = 0 \implies 3x_1 = 0 \implies x_1 = 0$.
  • Substitute $x_1 = 0$ into (3): $(0) + 3x_3 = 0 \implies 3x_3 = 0 \implies x_3 = 0$.
  • The variable $x_4$ does not appear in any equation (it is not a pivot variable), so it is a free variable. We can set it to any value. Let $x_4 = t$.

The general solution vector $\mathbf{x}$ is: $\mathbf{x} = \begin{pmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ t \end{pmatrix} = t \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}$

The null space is spanned by the single vector $(0, 0, 0, 1)$.

  • A basis for the null space is ${(0, 0, 0, 1)}$.
  • The nullity (dimension) is 1.

✅ Correct Option:

  • ${(0, 0, 0, 1)}$

❓ Question 8: Maximum Rank of AB

Let $A$ be a $4 \times 3$ matrix and $B$ be a $3 \times 4$ matrix. If the null spaces of $A$ and $B$ are 3 and 2 dimensional vector spaces, respectively, then what is the maximum possible rank of the matrix $AB$?

Core Concepts

  1. Rank-Nullity Theorem: For any $m \times n$ matrix $M$, $\text{rank}(M) + \text{nullity}(M) = n$ (the number of columns).
  2. Rank of a Product: $\text{rank}(AB) \le \min(\text{rank}(A), \text{rank}(B))$.
  3. Zero Matrix: A matrix with $\text{rank} = 0$ is the zero matrix. The product of a zero matrix with any other matrix (where defined) is the zero matrix.

Solution

  1. Find $\text{rank}(A)$:
    • $A$ is a $4 \times 3$ matrix ($n=3$).
    • We are given $\text{nullity}(A) = 3$.
    • Using Rank-Nullity: $\text{rank}(A) + \text{nullity}(A) = n$
    • $\text{rank}(A) + 3 = 3$
    • $\text{rank}(A) = 0$
  2. Find $\text{rank}(B)$:
    • $B$ is a $3 \times 4$ matrix ($n=4$).
    • We are given $\text{nullity}(B) = 2$.
    • Using Rank-Nullity: $\text{rank}(B) + \text{nullity}(B) = n$
    • $\text{rank}(B) + 2 = 4$
    • $\text{rank}(B) = 2$
  3. Find $\text{rank}(AB)$:
    • We know $\text{rank}(A) = 0$. A matrix with rank 0 is the zero matrix.
    • So, $A$ is the $4 \times 3$ zero matrix ($A = \mathbf{0}_{4 \times 3}$).
    • The product $AB$ is $A_{4 \times 3} B_{3 \times 4}$.
    • $AB = \mathbf{0}{4 \times 3} \cdot B{3 \times 4} = \mathbf{0}_{4 \times 4}$
    • The product $AB$ must be the $4 \times 4$ zero matrix.
    • The rank of the zero matrix is 0.
    • Therefore, the only possible rank for $AB$ is 0.

The maximum possible rank is 0.

(Self-check using the inequality: $\text{rank}(AB) \le \min(\text{rank}(A), \text{rank}(B)) \implies \text{rank}(AB) \le \min(0, 2) \implies \text{rank}(AB) \le 0$. Since rank cannot be negative, $\text{rank}(AB) = 0$.)

✅ Final Answer: 0

Here are the detailed solutions and concepts for each of the questions you provided.

❓ Question 9: Dimension of Rating Space

Choose the set of correct options from the following.

  • If Shubham and Poulami both gave 2-stars rating, Asrifa and Raphael both gave 1-star rating, and Vicky gave 3-stars rating, then the dimension of the vector space spanned by the row vectors corresponding to the ratings is 3.
  • If Shubham and Vicky both gave 3-stars rating, Poulami gave 2-stars rating, Asrifa gave 5-stars rating, and Raphael gave 1-star rating then the dimension of the vector space spanned by the row vectors corresponding to the ratings is 4.
  • If Shubham gave 5-stars rating, Poulami gave 3-stars rating, Asrifa gave 1-star rating, Raphael gave 2-stars rating, and Vicky gave 4-stars rating then the dimension of the vector space spanned by the row vectors corresponding to the ratings is 5.
  • If Shubham, Poulami, and Asrifa, all gave 2-stars rating, and Raphael and Vicky gave 1-star rating then the dimension of the vector space spanned by the row vectors corresponding to the ratings is 1.

Core Concept 🧠

The “dimension of the vector space spanned by the row vectors” is simply the rank of the matrix formed by those row vectors. The rank is the total number of linearly independent rows.

A 1-star rating is the vector [1 0 0 0 0]. A 2-star rating is [0 1 0 0 0]. A 3-star rating is [0 0 1 0 0]. A 4-star rating is [0 0 0 1 0]. A 5-star rating is [0 0 0 0 1]. No rating is [0 0 0 0 0].

These rating vectors are special: they are the standard basis vectors (plus the zero vector). They are, by definition, linearly independent of each other. Therefore, the rank is simply the number of unique ratings given (excluding the ’no rating’ vector).

Solution

  1. Check Option 1:

    • Ratings: 2-star, 2-star, 1-star, 1-star, 3-star.
    • Unique Ratings: {1-star, 2-star, 3-star}.
    • Vectors: [1 0 0 0 0], [0 1 0 0 0], [0 0 1 0 0].
    • These 3 vectors are linearly independent. The rank is 3.
    • This option is CORRECT. ✅

  2. Check Option 2:

    • Ratings: 3-star, 3-star, 2-star, 5-star, 1-star.
    • Unique Ratings: {1-star, 2-star, 3-star, 5-star}.
    • Vectors: [1 0 0 0 0], [0 1 0 0 0], [0 0 1 0 0], [0 0 0 0 1].
    • These 4 vectors are linearly independent. The rank is 4.
    • This option is CORRECT. ✅

  3. Check Option 3:

    • Ratings: 5-star, 3-star, 1-star, 2-star, 4-star.
    • Unique Ratings: {1-star, 2-star, 3-star, 4-star, 5-star}.
    • Vectors: [1 0 0 0 0], [0 1 0 0 0], [0 0 1 0 0], [0 0 0 1 0], [0 0 0 0 1].
    • These 5 vectors are the standard basis for $\mathbb{R}^5$ and are linearly independent. The rank is 5.
    • This option is CORRECT. ✅

  4. Check Option 4:

    • Ratings: 2-star, 2-star, 2-star, 1-star, 1-star.
    • Unique Ratings: {1-star, 2-star}.
    • Vectors: [1 0 0 0 0], [0 1 0 0 0].
    • These 2 vectors are linearly independent. The rank is 2.
    • The option says the dimension is 1. This option is INCORRECT. ❌

✅ Correct Options:

  • The first, second, and third options are all correct.

❓ Question 10: Analysis of Matrix A

Consider the matrix $A$ corresponding to the ratings given in Table M2W5G2, which is given as follows: $A = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \ 1 & 0 & 0 & 0 & 0 \ 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$

Which of the following statements about the matrix $A$ are true?

  • The set {(1, 0, 0, 0, 0), (0, 1, 0, 0, 0), (0, 0, 0, 1, 0)} spans the nullspace of $A$.
  • The set {(0, 0, 1, 0, 0), (0, 0, 0, 1, 0)} spans the nullspace of $A$.
  • Nullity of $A$ is 2.
  • Nullity of $A$ is 3.

Core Concepts 🧠

  1. Null Space: The null space of a matrix $A$ is the set of all vectors $\mathbf{x}$ that solve the homogeneous equation $A\mathbf{x} = \mathbf{0}$.
  2. Nullity: The nullity is the dimension of the null space. It’s the number of free variables in the solution to $A\mathbf{x} = \mathbf{0}$.
  3. Rank-Nullity Theorem: For any $m \times n$ matrix, $\text{rank}(A) + \text{nullity}(A) = n$ (where $n$ is the number of columns).

Solution

  1. Find the Null Space: We need to solve $A\mathbf{x} = \mathbf{0}$. Let $\mathbf{x} = \begin{pmatrix} x_1 \ x_2 \ x_3 \ x_4 \ x_5 \end{pmatrix}$. The system of equations is:

    • $0x_1 + 1x_2 + 0x_3 + 0x_4 + 0x_5 = 0 \implies x_2 = 0$
    • $1x_1 + 0x_2 + 0x_3 + 0x_4 + 0x_5 = 0 \implies x_1 = 0$
    • $1x_1 + 0x_2 + 0x_3 + 0x_4 + 0x_5 = 0 \implies x_1 = 0$ (This is a redundant equation)
    • $0x_1 + 0x_2 + 0x_3 + 0x_4 + 1x_5 = 0 \implies x_5 = 0$
    • $0 = 0$ (This is always true)

  2. Identify Variables:

    • Pivot Variables (solved for): $x_1, x_2, x_5$.
    • Free Variables (not solved for): $x_3, x_4$.
    • We have 2 free variables. This immediately tells us the nullity is 2.

  3. Find the Basis:

    • We know $x_1 = 0$, $x_2 = 0$, $x_5 = 0$.
    • Let $x_3 = s$ (a free parameter).
    • Let $x_4 = t$ (another free parameter).
    • The solution vector $\mathbf{x}$ is: $$\mathbf{x} = \begin{pmatrix} 0 \\ 0 \\ s \\ t \\ 0 \end{pmatrix} = s \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}$$
    • The null space is spanned by the set {(0, 0, 1, 0, 0), (0, 0, 0, 1, 0)}.

  4. Check the Options:

    • The set {(1, 0, 0, 0, 0), (0, 1, 0, 0, 0), (0, 0, 0, 1, 0)} spans the nullspace... False. This set is the basis for the row space (the space spanned by the rows).
    • The set {(0, 0, 1, 0, 0), (0, 0, 0, 1, 0)} spans the nullspace... True. This is exactly what we found.
    • Nullity of A is 2. True. We found 2 free variables.
    • Nullity of A is 3. False.

✅ Correct Options:

  • The set {(0, 0, 1, 0, 0), (0, 0, 0, 1, 0)} spans the nullspace of $A$.
  • Nullity of $A$ is 2.

❓ Question 11: Rank of Matrix A

What will be the rank of the matrix $A$ defined in the above question?

Core Concept 🧠

Rank is the number of linearly independent rows (or columns) in a matrix. There are several ways to find it:

  1. Count the number of non-zero rows in the RREF (Row Echelon Form).
  2. Count the number of pivot columns.
  3. By inspection, count the number of rows that cannot be created by adding/scaling other rows.
  4. Use the Rank-Nullity Theorem: $\text{rank}(A) = n - \text{nullity}(A)$.

Solution

Method 1: By Inspection

  • Row 1: [0 1 0 0 0]
  • Row 2: [1 0 0 0 0]
  • Row 3: [1 0 0 0 0] (This is identical to Row 2, so it’s linearly dependent)
  • Row 4: [0 0 0 0 1]
  • Row 5: [0 0 0 0 0] (This is the zero vector, always dependent) The unique, non-zero, linearly independent rows are (0, 1, 0, 0, 0), (1, 0, 0, 0, 0), and (0, 0, 0, 0, 1). There are 3 such rows. Therefore, the rank is 3.

Method 2: Using Rank-Nullity Theorem

  • From Question 10, we definitively found that $\text{nullity}(A) = 2$.
  • The matrix $A$ has $n = 5$ columns.
  • $\text{rank}(A) + \text{nullity}(A) = n$
  • $\text{rank}(A) + 2 = 5$
  • $\text{rank}(A) = 3$

Both methods give the same result.

✅ Final Answer: 3

❓ Question 12: Nullity Equivalents

Let $M$ be an $m \times n$ matrix and let $A$ be the reduced row echelon form (RREF) of $M$. Choose all the options which contain quantities equal to the nullity of $M$.

  • $m - \text{rank}(M)$
  • $n - \text{rank}(A)$
  • Number of non-zero rows in $M$
  • Number of pivot elements in $A$
  • Number of independent variables for the system of equations $Mx = 0$
  • Number of independent variables for the system of equations $Ax = 0$

Core Concepts 🧠

  1. Nullity: The dimension of the null space.
  2. Rank-Nullity Theorem: This is the key. $\text{rank}(M) + \text{nullity}(M) = n$ (number of columns). This can be rearranged to $\text{nullity}(M) = n - \text{rank}(M)$.
  3. Properties of RREF: Row reduction does not change the row space, the null space, or the rank.
    • $\text{rank}(M) = \text{rank}(A)$
    • $\text{nullity}(M) = \text{nullity}(A)$
    • The solution to $M\mathbf{x} = \mathbf{0}$ is the same as the solution to $A\mathbf{x} = \mathbf{0}$.
  4. Rank Definitions: $\text{rank}(A)$ = Number of pivot elements in $A$ = Number of non-zero rows in $A$.
  5. Nullity Definitions: $\text{nullity}(M)$ = Number of free variables in the system $M\mathbf{x} = \mathbf{0}$. (The term “independent variables” is often used to mean free variables, as their values can be chosen independently).

Solution

Let’s check each option against these facts.

  • $m - \text{rank}(M)$
    • The formula involves $n$ (columns), not $m$ (rows). Incorrect.
  • $n - \text{rank}(A)$
    • We know $\text{rank}(A) = \text{rank}(M)$.
    • So this expression is $n - \text{rank}(M)$.
    • By the Rank-Nullity Theorem, $\text{nullity}(M) = n - \text{rank}(M)$. Correct.
  • Number of non-zero rows in $M$
    • This is not a useful quantity. A matrix can have all non-zero rows and still have a large nullity (e.g., if all rows are identical). The important quantity is the number of non-zero rows in the RREF (A), which is the rank. Incorrect.
  • Number of pivot elements in $A$
    • This is the definition of $\text{rank}(A)$, which is $\text{rank}(M)$. This is the rank, not the nullity. Incorrect.
  • Number of independent variables for the system of equations $Mx = 0$
    • “Independent variables” (or free variables) are the variables not associated with a pivot column. The number of free variables is the dimension of the null space. This is the definition of nullity. Correct.
  • Number of independent variables for the system of equations $Ax = 0$
    • Since $A$ is the RREF of $M$, the systems $M\mathbf{x}=\mathbf{0}$ and $A\mathbf{x}=\mathbf{0}$ have the exact same solution set.
    • Therefore, they have the same number of free (independent) variables. This is also the nullity. Correct.

✅ Correct Options:

  • $n - \text{rank}(A)$
  • Number of independent variables for the system of equations $Mx = 0$
  • Number of independent variables for the system of equations $Ax = 0$

❓ Question 13: Non-Injective Linear Transformation

Let $T : \mathbb{R}^2 \to \mathbb{R}^2$ be a function for which the following is known: $T(1, 0) = (k, 1)$, $T(0, 1) = (1, k)$, $T(2, 1) = (-k^2 + k + 1, 2 - k^2)$, where $k \in \mathbb{R}$. Find the value of $k$ for which $T$ is a linear transformation which is not injective.

Core Concepts 🧠

This problem has two separate conditions that $k$ must satisfy:

  1. Condition for Linearity: If $T$ is linear, it must satisfy $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ and $T(c\mathbf{u}) = cT(\mathbf{u})$.
    • We can use $T(2, 1) = T(2(1,0) + (0,1)) = 2T(1,0) + T(0,1)$.
  2. Condition for “Not Injective”: For a square matrix $A$ representing a linear map $T$, “not injective” (not one-to-one) is equivalent to:
    • $A$ is not invertible.
    • $\det(A) = 0$.

We must find the value(s) of $k$ that satisfy Condition 1, find the value(s) that satisfy Condition 2, and then find the intersection of those two sets.

Solution

Step 1: Find $k$ that makes $T$ linear.

  • Let’s calculate what $T(2, 1)$ must be if $T$ is linear: $T(2, 1) = 2 \cdot T(1, 0) + 1 \cdot T(0, 1)$ $T(2, 1) = 2 \cdot (k, 1) + 1 \cdot (1, k)$ $T(2, 1) = (2k, 2) + (1, k)$ $T(2, 1) = (2k + 1, 2 + k)$
  • The problem gives us $T(2, 1) = (-k^2 + k + 1, 2 - k^2)$.
  • For $T$ to be linear, these two must be equal:
    • X-component: $2k + 1 = -k^2 + k + 1 \implies 2k = -k^2 + k \implies k^2 + k = 0 \implies k(k+1) = 0$.
    • Y-component: $2 + k = 2 - k^2 \implies k = -k^2 \implies k^2 + k = 0 \implies k(k+1) = 0$.
  • Both components give the same equation. The values of $k$ that make $T$ linear are $k=0$ or $k=-1$.
    • Linearity Set: $k \in {0, -1}$.

Step 2: Find $k$ that makes $T$ “not injective”.

  • First, we build the matrix $A$ for the transformation $T$. The columns of $A$ are the images of the basis vectors, $T(1,0)$ and $T(0,1)$.
    • $A = \begin{bmatrix} T(1,0) & T(0,1) \end{bmatrix} = \begin{bmatrix} k & 1 \ 1 & k \end{bmatrix}$
  • For $T$ to be not injective, the matrix $A$ must not be invertible, which means $\det(A) = 0$.
  • $\det(A) = (k \cdot k) - (1 \cdot 1) = k^2 - 1$.
  • Set the determinant to zero: $k^2 - 1 = 0 \implies (k-1)(k+1) = 0$.
  • The values of $k$ that make $T$ not injective are $k=1$ or $k=-1$.
    • Not Injective Set: $k \in {1, -1}$.

Step 3: Find the $k$ that satisfies BOTH conditions.

  • We need $k$ to be in the “Linearity Set” AND the “Not Injective Set”.
  • $\text{Intersection}({0, -1}, {1, -1}) = {-1}$
  • The only value that works is $k = -1$.

✅ Final Answer: -1

Graded Assignment 4Graded Assignment 6